Problem 12 - Precalculus: Functions and Graphs

Question

Solve the system. $$\left\\{\begin{array}{l} 0.11 x-0.03 y=0.25 \\ 0.12 x+0.05 y=0.70 \end{array}\right.$$

Step-by-Step Solution

Answer

The solution is approximately \(x \approx 3.68\) and \(y \approx 5.16\).

Step 1: Multiply equations to eliminate decimals

Multiply the first equation by 100 to eliminate the decimals, which gives us:\[11x - 3y = 25\] Similarly, multiply the second equation by 100:\[12x + 5y = 70\] Now we have a system of equations with whole numbers.

Step 2: Use elimination method to cancel out y

To eliminate \(y\), let's multiply the first equation by 5 and the second equation by 3, so their coefficients of \(y\) match:\[5(11x - 3y) = 5(25) \3(12x + 5y) = 3(70)\]This transforms our system to:\[55x - 15y = 125 \36x + 15y = 210\] Now, add the equations to eliminate \(y\).

Step 3: Solve for x

Add the equations from Step 2:\[55x - 15y + 36x + 15y = 125 + 210 \91x = 335\] Now, solve for \(x\) by dividing both sides by 91:\[x = \frac{335}{91} \approx 3.68\] So, \(x \approx 3.68\).

Step 4: Substitute x to find y

Use the value of \(x\) from Step 3 in the first equation (in whole number form):\[11(3.68) - 3y = 25\]Calculate:\[40.48 - 3y = 25\]Rearrange to find \(y\):\[-3y = 25 - 40.48 \y = \frac{-15.48}{-3} \approx 5.16\] Thus, \(y \approx 5.16\).

Step 5: Verify solution

Check the values of \(x\) and \(y\) by substituting them back into the original equations:For the first equation:\[0.11(3.68) - 0.03(5.16) \approx 0.25\]And for the second equation:\[0.12(3.68) + 0.05(5.16) \approx 0.70\]Both equations hold true, thus the solution is verified.

Key Concepts

Elimination Method

The Elimination Method is a classic technique used to solve a system of linear equations. It involves adding or subtracting equations to eliminate one of the variables, making it easier to solve for the remaining variable.

Here's how it works:

• You adjust the coefficients of one of the variables in the system so that they are opposites, effectively canceling each other out when the equations are added or subtracted.
• Once a variable is eliminated, you solve for the remaining variable.
• Then, substitute this value back into one of the original equations to find the other variable.

This method is especially useful when the coefficients make elimination straightforward, or when systems are more complicated than simple substitution would allow.

System of Linear Equations

A System of Linear Equations consists of two or more linear equations involving the same set of variables. The goal is to find a common solution where all equations intersect.

For example, in the exercise given:

• The system includes two equations: 1\. \(0.11x - 0.03y = 0.25\) 2\. \(0.12x + 0.05y = 0.70\)
• Each equation represents a line on a graph, and the solution to the system is the point where these lines intersect.

Different methods such as elimination, substitution, or graphical methods can be used to find the solution to such systems. The method chosen usually depends on the specific details of the problem and personal preference or ease.

Decimal Elimination

Decimal Elimination is a handy technique used when linear equations include decimal numbers. By transforming equations into whole numbers, calculations become more manageable.

Here's how you perform decimal elimination:

• Identify the smallest power of 10 that makes all the decimals in the equation whole numbers.
• In this problem, multiplying each term by 100 gets rid of two decimal places, giving whole number coefficients: - First Equation: \(0.11x - 0.03y = 0.25 \) becomes \( 11x - 3y = 25 \) - Second Equation: \(0.12x + 0.05y = 0.70 \) becomes \(12x + 5y = 70 \)

This simplification makes it easier to apply other solving methods like elimination or substitution, avoiding errors that can arise from working with decimals.